求[H]ea: [HAclea= 5.6×1010 1.8×10 ×0.01×0.01 =5.6×10- [H1= K.[HAcl_18×103×5.6×10 =1.0×107 [Acle 0.01 pH=7.00 TE= D1(10-10p) △pH=t0.2 CHAc.q TE= 56×10-5×(102-10-02 =±0.005=±05% 0.01 或TE= 2[D sinh(2303△pH) CHAc.eq 2×5.6×10 sinh(2303×02)=±5×10 0.01求[H + ]eq： pH 7.00 1.0 10 0.01 1.8 10 5.6 10 [Ac ] [HAc] [ H ] 0.01 0.01 5.6 10 1.8 10 5.6 10 [HAc ] 7 5 5 eq a eq eq 5 2 1 5 10 eq = = × × × × = = = ×         × × × × = − − − − + − − − K HAc,eq ∆pH ∆pH eq [D] (10 10 ) TE c − − = ∆pH = ± 0.2 3 5 HAc,eq eq 5 0.2 0.2 sinh( 2.303 0.2 ) 5 10 0.01 2 5.6 10 sinh( 2.303 pH ) 2[D] TE 0.005 0.5 % 0.01 5.6 10 (10 10 ) TE − − − − × = ± × × × = ± = ∆ = ± = ± × × − = ± c 或